document.write( "Question 207022: A recent study showed that 75 % of commercial airline flights in and out of the US airports were on-time arrivals and 19 % were late departures. three hundred flights are to be randomly identified from all flights and their flight logs examines closely. \r
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Algebra.Com's Answer #156530 by Edwin McCravy(20056) $\"\"$ $\"About$

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A recent study showed that 75 % of commercial airline flights in and out of the US airports were on-time arrivals and 19 % were late departures. three hundred flights are to be randomly identified from all flights and their flight logs examines closely.
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document.write( "This is a binomial probablity, where\r\n" );
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document.write( " = 80% of 300 = \r\n" );
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document.write( "If , then this is approximately normally\r\n" );
document.write( "distributed.  , so\r\n" );
document.write( "the normal approximation will be close.\r\n" );
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document.write( "Since we read the words \"more then 80 percent\", we do not\r\n" );
document.write( "include the 240 bar of the implied histogram, the base of\r\n" );
document.write( "which goes from 239.5 to 240.5.  So the first bar of the\r\n" );
document.write( "histogram we include starts at 240.5. So we take that value\r\n" );
document.write( "for x:  \r\n" );
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document.write( ", rounded to \r\n" );
document.write( "nearest hundredth.\r\n" );
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document.write( "Now we need to find out the area under the normal curve\r\n" );
document.write( "to the right of the black vertical line at 2.07\r\n" );
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document.write( "Normal tables are set up to give the area between the y-axis and\r\n" );
document.write( "that black line drawn at . \r\n" );
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document.write( "So we look down the left column, till we see 2.0,then across the top\r\n" );
document.write( "to the hundredth digit, which is .07, so we read .4808.  Since all the\r\n" );
document.write( "area to the right of the y-axis is .5, we subtract .5-.4808 and\r\n" );
document.write( "get .0192.\r\n" );
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document.write( "Edwin

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